To determine how Philip can sort the acorns in equal groups with the same number of tops and bottoms, we need to look for a factorization that splits the total of 92 acorns evenly into groups, while balancing the number of acorns with and without tops.
Philip has:
- 66 acorns with tops
- 26 acorns without tops
To satisfy his requirement of equal grouping, we can use the greatest common divisor (GCD) or other factors of the total acorns to create groups.
Let's check the options:
-
2(33 + 13):
- This represents 2 groups with 33 acorns with tops and 13 acorns without tops in each group.
- Checking: 2 groups mean \(2 \times 33 = 66\) with tops and \(2 \times 13 = 26\) without tops, which is correct.
-
6(11 + 4):
- This represents 6 groups with 11 acorns with tops and 4 acorns without tops in each group.
- Checking: 6 groups mean \(6 \times 11 = 66\) with tops and \(6 \times 4 = 24\) without tops, given that 24 is not adequate since he has 26 without tops.
-
1(66 + 26):
- This represents 1 group with all acorns. This does not fit the requirement of sorting into groups.
-
24(3 + 1):
- This represents 24 groups with 3 tops and 1 without in each group.
- However, checking shows \(24 \times 3 = 72\) with tops, which exceeds the number he has.
The best representation of how Philip can sort the acorns into equal groups, maintaining the balance of acorns with and without tops, is 2(33 + 13).
Final answer: 2(33+13)
1 answer